 Zato, čestnji sprijev z konferenci. Alexi Figali, kaj je ga dobročil? Zvuknil, da je zvuknil in da je vsega vsega in vsega. Zvuknil, da je vsega in vsega in vsega. Tako, da so način. Zvuknil, da je in vsega in vsega. So, da je zvuknil. So, da je zvuknil, da je zvuknil. The last time was clouding. So, what I would like to present in these two hours is some recent result about the georgic conjecture. So, I will spend essentially the first hour to give an introduction to the problem. We discussed the classical georgic conjecture for the classical applazion, what are the non-results. Then I will conclude this talk with the statement of the theorem that we recently proved, and then in the second part of the lecture I will discuss more the technical aspects. So, thanks for being here still at the end of the last of the two weeks. Consider also that today is a national holiday in Italy, so, especially for the Italians present, present here it's even more challenging. So, let me start with the georgic conjecture. So, the georgic conjecture is the following. So, this is a conjecture that was formulated in 1978 and concerns solution to the following equation. So, I have a function u in rd, with values in 2r. u is bounded, let's say u between bounded by 1, and u solves the equation minus laplacian u equal u minus u cube. So, this is called Allen-Kan equation. So, it's a global solution in the workspace, it's bounded, and you assume in addition one extra hypothesis, which is assume that u is monotone in some direction, let's say in the last direction, in the last variable, that is the partial derivative with respect to the last variable of u is trity positive. So, you have a function in rd, this is rd minus 1, this is the last coordinate, and you assume that your function increase in this variable. I don't know how to do it, but as you move up, the value here is just increasing in each vertical segment. Then, if d is less than or equal to 8, u is one dimension, that is, u of x can be written as a function phi of x dot sigma, where sigma is a vector on the unit sphere, and phi is a solution, is a 1D solution to the same problem. So, it's just a solution to the od minus phi second equal phi minus phi cube. So, this is the Georgi conjecture, and what I want to do is to explain why, how the Georgi reached this conclusion, so how he came up with this. Ok, so to explain this, we have to make a comparison between all in kan versus minimal surfaces. Ok, so u soluša of all in kan is equivalent to say, so this is the Euler Lagrange equation for some energy function. So, it's equivalent to say that u is a critical point for the following energy, i1 over function v is the integral of grad v square over 2 plus 1 minus v square over 4. Ok, so, you can check, if you want to exercise, you check that if you do the partial derivative with, I mean, you differentiate with the epsilon, the energy of u plus epsilon w, and then this is after integration by parts, you get exactly minus, let's see, Laplace nu plus u minus u cube times w. So, you see that u is a soluša, so if this is a critical point means that this is 0 for every w, and the only way this can be 0 for every w is that this is 0, which means you are a soluša of all in kan, and vice versa, if you are a soluša of all in kan, you are a critical point. So, you get this. So, soluša to all in kan are critical points for this energy. Now, let me make an extra assumption to make this connection. Let's assume that we are dealing with minimizers. So, let's assume, so now I'm just doing a series of heuristic arguments to explain how you come up with this conjecture. So, now we're gonna put extra assumptions on my solution. So, let's assume that u is a local minimizer, that is, what do I mean by local minimizer? It just means local not because it's a local for small perturbation, local because it's a minimizer for comparatively supported perturbations. That is that the energy of u plus w, let's say, is always minus the energy of u is always greater or equal to zero for every w, which is, let's say, c1, and comparatively supported in rd. You see, I have to write inequality in this way. I cannot write these energies greater than these energies because in general, maybe both quantities are infinite. I mean, this could be both infinite. And when you've write the difference, because this is comparatively supported, you can only, this is the same as the integral over a ball of radius r of grad u plus grad w square over 2 plus 1 minus u plus w square over 4. What's going on with this? I think I broke it when it felt changed. Minus grad u square over 2 minus 1 minus u square over 4 for any r such that the support of w is contained in br. Because outside br, these quantities are the same. So I just cancel out the integral outside, and now this is a finite integral. If I was integral over a random, maybe I get plus infinity. I get infinity. It's like if you take the area of an hyperplane, right? The area of an hyperplane is infinity. So it doesn't make much sense to say that hyperplane minimizes something. They minimize in compact sense. OK, so let's assume that u is a local minimizer in this sense. Then, and this is, you can check it, that u epsilon over x defined as u over x over epsilon is a local minimizer. This is not my fault. This is a local minimizer for a new energy, i epsilon of v, which is, you have the same energies there, but with some scanning. So now you have the integral of epsilon, grad v square over 2 plus 1 over epsilon, 1 minus v square over 4. One can do the computation. You just need, you need to compare the energy of your epsilon with anything. I mean, you just use a minimizer. I don't need this exercise. Let me know if you have problems. So this is just any change of light. Anyhow, so u epsilon is a minimizer of this energy, and now the question is what can I infer from the minimality of this energy. So this energy, for epsilon small, this energy is kind of interesting, because it has two terms competing. One term that tells me that u epsilon square, because u epsilon is a minimizer, so u epsilon wants to make this quantity as small as possible. Here you see you have a very singular coefficient in front, one of epsilon is going to infinity, so you want to make this as small as possible, which means that u epsilon square should be close to 1, which means that a minimizer u epsilon should be close to plus and minus 1. So if you get plus and minus 1 here, you just make it 0. Now, what happens here? Here, every time, let's say, I have here u epsilon close to 1, and here I have an interface, let's say, a strip, here is close to 1, and here u epsilon is close to minus 1, here there is a transition, because you have to pass from one to another, and you're going to pay some gradient, and then this term is going to affect. OK, so we have a balance. One term is going to force my function to take values essentially only 1 and minus 1. The second term tells me, every time you jump from 1 to minus 1, you pay something, because there is a gradient penalization. OK, let's try to see if there is a bit more than you can say. So the philosophy is that we want to change, so whenever we change value, we pay, so maybe this part of the energy can tell me that minimizers are not arbitrary functions that go from 1 to minus 1. Maybe this interface, this region where I go from 1 to minus 1 has some special property. So let's try to do this. Let me do like an example. Let me do the following. I minimize the energy, I epsilon v, now in a domain omega, let's say, I have omega, among all functions v, from omega into R, let's say omega from minus 1, 1, I want to work in that region, such that the integral of omega is 0. I put 0 average. Why I want to do this, because if you just minimize this energy without any constraint, there are two trivial minimizers of this energy. You just take u equal 1, it has 0 energy, u equal minus 1, it has 0 energy. It makes this 0 and this 0. So you want to avoid trivial minimizers. But if I put the 0 average constraint, it means that there is 50% of the time I have to be positive, 50% of the time I have to be negative, I cannot be all the time 1 or minus 1. I have to have some interface somewhere. I have a picture. Let's say that u epsilon is close to 1 here. So let me call u epsilon again the minimizer, u epsilon is close to minus 1 here. And I have some interface, sigma, so sigma would be like the level set u epsilon equals 0, let's say. And then I ask myself, can I say something about sigma? So if assuming that u epsilon are minimizers. OK, so let's try to understand. Assume that let's call delta, let's call delta the width of the band, let's say this area along which the transition course. So here I think that u epsilon is super close to 1. It has no gradient and essentially no gradient of energy from both ends. Here is super close to minus 1. Here is really where it transition. OK, so the picture is something like this and then it does this. This is the profile of u epsilon. So this is delta interval. Let's look now at this energy for u epsilon. When we look at the energy, so we have u i epsilon of u epsilon. This is going to be let me call maybe this interface. This region is omega delta is the transition. Omega epsilon, let's call it omega epsilon region where u epsilon transition. So the energy essentially is all concentrated in that strip because outside of omega epsilon the function is close to 1 and minus 1. So it doesn't pay anything in gradient, it doesn't pay anything in the second term. So all the energy lives in omega epsilon. So this is essentially the integral over omega epsilon of the gradient of u epsilon square over 2 with an epsilon in front plus 1 over 4 epsilon 1 minus u epsilon square square. OK, now let's try to understand this. Omega epsilon is a region like this. I think it's a very smooth, nice region. Delta is the strip. This is my surface sigma. So if my function has to go from 1 to minus 1 in a region of length delta here the gradient of u epsilon is going to be inside 1 over delta. 2 over delta. I have to go from 1 to minus 1 and length delta is 2 over delta is the slope of my function. I will forget constants and this doesn't matter. So here, what do I get? Epsilon up to constants. The energy is essentially epsilon. Then I get the gradient of this which is 1 over delta square. So delta square. And then the volume of omega epsilon. And for the second one I pay here. So when I'm here in between my function is far from 1 to minus 1 here my function is between 1 half and minus 1 half. So here this is over the 1 is not 0. It's some quantity. Some positive constant. So here I pay 1. I have 1 over epsilon then I have the volume. So I have 1 over epsilon the volume of omega epsilon. How much is this volume? The volume of this strip is delta, the width times the area of the surface. So this is essentially epsilon over delta square. And then I have delta coming from here. So this is volume of epsilon is essentially delta times the area I will use it to the hd minus 1 of sigma. So this is the dimensional surface area of sigma. So delta simplifies. I have epsilon over delta plus delta over epsilon hn minus 1 of sigma. So this tells me, so my epsilon should minimize this quantity essentially. Because that's the energy. So to minimize this I want to take so what is the best delta here? Delta of the freedom. Minimizing delta is delta square epsilon. So you take delta equal epsilon that's the minimization of this. And then I take sigma to be a surface that minimizes the area that's the surface area. So among all possible surfaces sigma that I can think of I want to take a surface that minimizes the area among all surfaces in two walls of equal volume. Why this? Because you have omega here here you have sigma, let's say and you see here your epsilon is close to 1 here your epsilon is close to minus 1. Let's say this is omega 1 and this is omega 2, the two pieces is these two regions. So the integral of omega of your epsilon by summation is 0. But then this is essentially the volume of omega 1 because here is 1, minus the volume of omega 2 because this is minus 1. So the average zero constraint tells me that essentially the volume of these two pieces should be the same up to arrows in epsilon. So the sigma has to cut the domain in two walls of the same volume. These come from the non-triviality constraint so this argument is telling me essentially how I expect to be your epsilon, your epsilon for epsilon very small should be a function which is 1 and then goes from 1 to minus 1 and the interface the transition occurs along a surface which minimizes the area among all possible surfaces with this half volume constraint. So this is a minimal surface constraint I told you because I wanted the problem to be non-trivial. But so here we see minimal surface appearing. The minimal surface appear is essentially the limit of the level set so sigma is the level set your epsilon equals 0 because they are taking your epsilon equal 1 half your epsilon equals minus 1 half it doesn't matter because as epsilon goes to 0 the level sets collapse there. So any level set between minus 1 and 1 works. Let's say your epsilon equals 0 just by symmetry. Ok, so so we got an extra information from the minimality. Not only the function goes from minus 1 to 1 but the transition the level set your epsilon equals 0 should be able like a minimal surface. Is this clear? So this is, I think, the key point because this is where the dimension 8 and everything will come. Because now we found a connection between all and can and minimal surfaces. So let me give you just a precise statement that I will not discuss in the sense that it's just for people who are already familiar with gamma convergence but there is a theorem that is just making rigorous, let's say informal argument so Modica and Mortola one year before the conjecture 77 proved that as epsilon tends to 0 the energy I epsilon gamma converge to the perimeter function. So this is just a statement that makes rigorous what I said if you know gamma convergence well, if you don't know I will not explain to you because anyhow it will be completely relevant for me. What matters to me is just the corollary of this that as epsilon tends to 0 my u epsilon will converge let's say in l1 log to some function which is 1 and minus 1 so it's a characteristic of a set e minus the characteristic of rd minus e so it's a function which takes only values plus and minus 1 so here is e so here I have 1, here I have minus 1 this function and e is a local minimizer of the perimeter that is the perimeter of f minus the perimeter of e is not negative for every f such that f outside of some ball is equal to outside of some ball for some r this is the analog of compactly supported perturbation so what I am saying is that do f converge work this is white what I am saying is that e is a minimality property and the minimality is the form you take a ball and this is not the best you take a ball and you take any function any set f which is equal to the outside let's say this is f so f is this set and then for every set f like this perimeter so this area the area of the interface here between f and the complement is larger than the area of f of the area of e so this is a regular statement what I did was just heuristic and this is the theorem ok so now probably move there we want to understand the structure of minimal surfaces because if we understand the structure of minimal surfaces so sets that minimize the perimeter like that I should be careful with the terminology let's call minimal surfaces sometimes are not minimizers so if I understand the structure of local minimizers of the perimeter then I will get information on my U epsilon for epsilon very small so let's discuss these local minimizers of the perimeter so the classification of local minimizers of the perimeter so first result e local minimizer of perimeter of perimeter in rd d less or equal to 7 then e is a half space so there are no local minimizers of the perimeter in dimension 7 besides the trivial ones so you have a set in volar rd and you assume that that property is true then it must be a plane second if d equal 8 the set e define in this way you take the couple x and y in r4 cross r4 such that the normal x is greater than the normal y is a local minimizer of perimeter so dimension 8 you take r4 r4 you take this set this set is a minimizer and it's not a half plane ok so here we get already some appearance of this number but in fact these dozen times were really wide dimension 8 because it looks here that things are good up to dimension 7 and they are bad from dimension 8 on while my conjecture is up to dimension 8 so there is one information I'm missing so remark the assumption so in the georgie in the georgie u satisfies that the partial derivative of u in the last coordinate is positive so you have a function here u which is increasing so if I take let's say u equal 0 the level set u equal 0 this should be essentially a graph with respect to this last coordinate because it's monoton so in each direction you can only intersect in one point so because of that this set can never converge to this set which is not at all a graph so you see the counter example I gave it to you in dimension 8 it's not really compatible with that picture because there is an extra information that I'm not using here which is the level set essentially graph with respect to one variable so u satisfies that means that u equal 0 is a graph this is not really true because a priori u equal 0 could go to infinity before it's not a global graph maybe but ok, is a graph define some domain is a graph of with respect to rd minus 1 so I don't care only on minimal surface I care about minimal surface that are also graph and so this new property e local minimizer plus e is a graph so let's say e is equal to the set of couples of the form x prime xd where xd is larger than h of x prime so is an epi graph so e is above the graph of the function for some h from rd minus 1 to r so if you assume this extra assumption that e is an epi graph and d now is less than 8 then e is a half space 2b d equal 9 is false so there is a counter example in dimension 9 I will not write it down because there is not such a clean nice formula but you see that we gain one dimension extra because of this graph property ok so now you start to see things patching together we got the dimension 8 and we got a connection to minimal surfaces so let's try now to conclude this discussion which motivates the georgi contract what is all this argument telling us this is for u epsilon so we took u, we consider u epsilon and we found these minimal graphs so the picture you should have in mind is this so I have rd minus 1 I have r and here I have my function u equal 0 let's say this is u equal lambda positive and so on these are the level sets of u and now what you do you do what is called a blow down so you shrink the picture here a ball of radius r and I shrink it down to a ball of radius epsilon r and whatever was here I shrink it here so now I have this new picture here and this now become the level sets of u epsilon equal lambda so u epsilon equal lambda by the function of u epsilon is equivalent to take u equal lambda and shrink it by factor epsilon and now so these are still global because these are global function and now as epsilon tends to 0 if the dimension is less than 8 this function should converge to a minimizer of the perimeter which we know to be only all space by this by 1 bis so here I use 1 bis gamma convergence result this tells me that if epsilon goes to 0 I should see plus 1 minus 1 so essentially all the level sets become shrink more and more so my function here is very u epsilon here is super close to 1 u epsilon here is super close to minus 1 shrink down and then in the limit I just see this transition and the transition is exactly on hyper plane so this is the picture and then what the Georgian conjecture is that the only way for this picture to happen is in fact that this level so the only way this level sets can be close to half space is to hyper plane to this flat if they were already hyper plane so the Georgian conjecture is this the Georgian said so we know epsilon equal lambda is close to hyper plane for epsilon small then indeed it was hyper plane to start with so for every, this is for every lambda in minus 1 wallet so you take any level set you know that in the limit they converge to hyper plane then he said well maybe they were already hyper plane and if they were hyper plane it means I have a function in rd whose level sets are hyper plane the function depends only on one variable which is the variable orthogonal to the hyper plane so if the level sets now are hyper plane this is now epsilon equal lambda well this implies that u epsilon depends only on one variable depends only on the orthogonal direction so u epsilon is 1d but u epsilon 1d means u is 1d from u to u epsilon is just a change of variable so if epsilon is small enough u epsilon is 1d then u is 1d is this clear? so the Georgian conjecture is motivated by this important remark all this argument works under the assumption so all this heuristic is based on the fact that u is a minimizer have this modica mortola and then blah blah here the Georgian never says that u is a minimizer so remark heuristic is based on the extra assumption u minimizer but the Georgian does not assume so in fact the Georgian is expected more than that expected that the result will be true just for monotone critical points but so somehow you can the argument you sketched you can really complete it on local minimizers you can make a proof out of it well that's what I will say it in a second sub indeed but it's difficult I mean you have to prove that you are flat enough then you are really flat I agree with that the last step is but I was just to check so it's true you can complete the argument make it rigorous that's what took essentially 30 years from the moment the Georgi say to the moment of video saving found a proof so it's a very non trivial proof but yeah you can complete that argument you have to prove that you are flat enough your release a hyper plane there is a gap so in fact let me state the results about the Georgi so when you are in dimension 2 there has been a proof by the sub in 98 they proved the Georgi conjecture then d equal 3 this was solved by Ambrosio and Cabre in 2000 then if you go from dimension 4 to 8 7 does it assuming u if you take the function u and you take the limit u x prime xd so you look at the function u and you take the limit when xd equals to plus and minus infinity so xd converge to plus and minus infinity then the limit is plus and minus 1 so we assume that when you go up in the limit you get 1 when you go down in the limit you get minus 1 and let me mention that this assumption so this assumption implies u is a local minimizer so it's stronger than assuming a local minimizer so it's not like a completely innocent assumption it's not obvious but it's true so essentially the result in dimension 4 to 8 is open up to now unless you assume this extra assumption is it equivalent to be a local minimizer or is it not clear I would think if you are a local minimizer is it equivalent to be a local minimizer a posteriori I think one can prove the equivalence if it's equivalent to be a local minimizer I think a posteriori yes up to dimension 8 I think in the posteriori yes I would say no no no it's just the assumption instead of assuming this can you just assume local minimizer and can you prove the result I would say yes up to dimension monotonizi plus local minimality the argument would work it's not exactly as it does it but it would work I think there is an extra case to handle if you don't assume that he knows out of this assumption that the hyperplane is like this I think without that assumption you may risk that your hyperplane kills but it would work I would bet it would work because it doesn't need it essentially it's there but let's say yes it is this way but if I had too bad I should check but if I had too bad I would say yeah it's enough local minimality and then dimension 9 dimension 9 delpino kovalčik kovalčik kovalčik c, z, y, k c, z, y, k and wait sorry savin I didn't say it here so savin was 2009 2009 and then delpino, kovalčik and wait in 11 prove that is false so they built a counter example but not to the conjection because the georgie was smart guy so it never conjectionated in dimension 9 so it's not that it is proved it's just that they showed the conjection is sharp in dimension 9 you cannot prove the result and the counter example is built on the counter example for minimal surfaces so it's built out of this so this is the status of the art the state of the art for this problem but before going to the one I'm interested in the alfalaplacian which comes from boundary reactions let me make also some extra comments so monotone versus stable solutions ok so there are three facts which are true so fact one the fact that u is monotone implies that u is stable so what does it mean in stable stable is like saying that infinitesimally you are a minimizer so stable means that you take the energy of u plus epsilon w and you do the second derivative with respect to epsilon epsilon equals zero then this is non negative for every w c1 and compact support ok so out of this monotone you have some kind of minimizacion property but it's only infinitesimal it's much less than say that you're a minimizer ok fact two if you define so because u is monotone you can define u plus and minus of x1 xd minus 1 to be the limit as xd tends to plus or minus infinity of u of x1 xd is the limit of your function at plus and minus infinity this is what for saving these are plus and minus 1 but in general there could be different functions and then you can prove u plus and minus are stable solutions of alenkan in one dimension less so you have a familiar solution of alenkan you take this limit in the last variable you get still solutions depending on one less variable and they are still stable not trivial but it's not too complicated you converge in a very smooth because these are all solutions to a pda when you take a translation of u you get a lot of compactness I mean these are very smooth to these functions and then fact three if u plus and minus are one dimension then u is one dimension so this is not obviously there but so this stream formation tells me something I could take the georgian conjecture the solution to the georgian problem define this limit and then try to say now I have a function defined in one dimension less I lost the monotonicity information I only keep the stability information and I still prove that the solutions are 1D notice that if I can prove that they are 1D then I get that u is 1D so these these three pieces tells me that if I can prove that these are 1D I get this but the advantage is that the functions u plus and minus are defined in one dimension less so the consequence the corollary would be to say that is that stable solutions are 1D in Rd minus 1 then monotone solutions are 1D in Rd so it means that if you can prove the result for stable in up to dimension 7 then you get the result for monotop to dimension 8 so this is if you want a stronger version of the georgian conjecture a more difficult and this is related to another conjecture if you are interested so this is related to the following conjecture if you take a surface so if you have boundary of E so E is a stable critical point for well in fact it is more general but boundary of E is a stable critical point for the perimeter then E is a hyper plane boundary of E is a hyper plane so what I am asking I am asking not that it is a minimizer but I am saying take a set E and assume that now you deform it a bit infinitesimally so you have E epsilon a smooth deformation of E and you assume that D over the epsilon of the perimeter of E epsilon epsilon equals 0 is 0 so this is saying that a critical point but also that it is stable and the derivative of the perimeter of E epsilon is greater or equal to 0 so it is not a global minimizer I don't have any information on the global picture I just know that infinitesimally first order cancels second is on negative then it should be an hyper plane if D is less or equal to 0 so this is the the analog this has been solved for D equal 3 by Fischer Colbre ah ja Colbre I screwed it and Richard in 80 and it was also do karma and Peng 79 and it is open in dimension 4, 5, 6 and 7 so if you want so you understand that going to stable with respect to minimizers is a big gap because even at the level of minimal surfaces what is known to be true for minimizers is still open for all the dimension between 4 and 7 so the stability word is delicate because you lose completely global information you are not allowed to deform your set in a very large ball or anything you just have infinitesimal ok, so let's try now just to go to the new problem let's see where we arrive then we take a break that's the good of having two hours don't have to rush ok, so the problem I want to study is a boundary reaction problem essentially you have deals so these problems come for many models from dislocation from boundary vortices I will not take I want to spend time on the modeling but let's say the problem is the following you just have so instead of minimizing like the the energy related to alenkan you have rd one dimension less r plus so you live in this hull space so I will call this rd plus 1 plus this is the hull space but you want to look so you want to understand so study local minimizers of the following energy integral over rd plus 1 of the gradient of v square over 2 here plus some boundary term integral over rd over xd plus 1 equal 0 so this is xd plus 1 of f of v where f is some function from r to r some boundary term so essentially you have inside the glasija delišli problem plus some boundary term in the energy ok and so you want to minimize this now the point is that you can do when you have a minimization like this you can look at it as a double minimization problem you can say well you see this second part of the energy only depends on the behavior of v here and then I have another part of the energy which depends on v here so what I can do is that I say ok I fix v on the boundary this part of the energy will not change and then I minimize this part of the energy among all functions that take the same value here and then I minimize the second part so this minimization over all v is the same as minimize among all functions little v from rd into r and then minimize among all functions w in rd plus 1 into r such that v is of x0 equal little v of x so here is x this energy and the energy is going to be grad v2 grad capital v2 over 2 over rd plus 1 plus and then the second energy only depends on f of little v because only depends on v on the lower dimension so here you see I have a double minimization but this part doesn't depend on capital v so I can first worry about this part and minimizing this energy with boundary condition is just the usual just get the harmonic extension minimizer is going to be harmonic is the harmonic function that takes this value so let's look at the minimizer of this so v minimizer minimizer implies that the Laplacian with respect to the old variable of v is equal to 0 and now let's look at this equation so this is I can look at this, this is what this is the Laplacian with respect to the x of v plus dxd plus 1 dxd plus 1 of v I wrote the Laplacian as the sum of the Laplacian with respect to the first coordinates and then the last now let me do a Fourier transform in this variable index so let v hat of psi xd plus 1 so I just do Fourier transform in the first variables so this is the integral of e i x c v xd plus 1 dx ok why I like it, because then in Fourier transform this Laplacian becomes very simple so then what you get so this part doesn't change because I didn't do anything in the last variable what you get this equation becomes dxd plus 1 dxd plus 1 of v equal minus sorry, no, the Laplacian is minus here you put it on this side is psi square v v hat, of course thank you so now we have a function of x of psi and xd plus 1 so you have now two variables e raise xd plus 1 and what I'm solving here for every fixed psi if you look just at this half line I'm just solving od, because I'm just taking second derivative v hat equal a constant times v hat and the solution I express it the solution of this is exponential so this tells me that v hat of psi xd plus 1 should be equal to the initial condition so v hat psi 0 times e to the minus mod of psi times xd plus 1 then you would say, ok, but wait the solution turns out to be there are two solutions there is the positive exponential the negative exponential but the positive exponential is blowing up while you know that your function when you take a nice function you take the harmonic extension this is a decaying function so there are two solutions but there is no so just by regularity you say, well, a priori there are two solutions but I can discard one of the two the one with the positive sign this cannot be so there is only the negative one so this is my solution, it's explicit and now let me rewrite the energy so what is the energy there the integral of a minimizer is grad w squared over 2 over rd plus 1 plus this I note that this is the same as the integral over rd plus 1 plus of one-half the divergence of grad v times v because you see the divergence of grad v times v so when you take the divergence of a product this is the divergence either follows on the first one give you laplacian v times v so when the divergence goes on the second and you get grad v scalar grad v so grad v squared and because v is harmonic the first term is zero so the grad v squared can be rewritten as other versions because I'm harmonic and then I can just do integration by parts so this is the integral one-half I mean the whole space the function is nice in the case x d plus 1 equals zero of v times so here is the normal derivative I will make the sign wrong so this should be minus the derivative respect to x d plus 1 of v is decreasing with a minus I think it's okay so this would be by the divergence theorem is the normal derivative but the normal derivative is so this is the normal derivative of v and the normal derivative v is minus the derivative in the x d plus 1 so that's why they put a minus okay but now I do in Fourier I say this is the same as one-half minus the integral of v at dx so this is again x d plus 1 equals zero dx d plus 1 of v at isometry in arc 2 so I keep this for x d equal to zero then what is the derivative I need this but I can compute it it's super simple because I have the formula so the derivative with respect to x d plus 1 of v at of xi x d plus 1 for x d plus 1 equals zero so you take the derivative you get when you differentiate this potential so you get minus modulus of xi v at of xi zero and then you have to take this for x d plus 1 equals zero which gives one so this is exponential minus modulus of xi x d plus 1 x d plus 1 equals zero so this is one so we found that the derivative is the function itself time minus model of xi so you substitute here minus and minus cancel of the integral over x d plus 1 equals zero of v at xi and v at again so v at square because I get v at and v at now I finish now just one last comment remember that v of x at zero is little v of x by assumption little v of x my definition was my trace my devalue xi zero is little v at of xi they have the same Fourier transform and so this for xi equals zero is nothing else that one half the integral over r d of xi v at square and for your information if you are used to that and I will discuss this later this is nothing else that one half the h one half norm h one half if you want you can invert this Fourier transform and this is one half integral double integral v of x minus v of y square divided by x minus y to the d plus 1 because the Fourier transform of xi is this so if you use that Fourier transform of xi is 1 over x to the d plus 1 you want the Fourier transform and you get convolution with this kernel so I stop here but now at least you saw that the problem was a boundary term problem and now I made this disappear and my minimization problem has become minimizing the h one half norm square plus f of v so you see now the alpha plane is coming because the alpha plane is the Euler Lagrange term that you come when you do the first variation of h one half so I stop here